Answer to Question #241984 in Physics for Kristine

Question #241984

The position of a squirrel running in a park is given by π‘Ÿ = [(310.280π‘š/𝑠)𝑑 + (0.0360π‘š/𝑠 2 )𝑑 2 ]𝑖̂+ (0.0190π‘š/𝑠 3 )𝑑 3 𝑗̂. (a) What are 𝑣π‘₯(𝑑) and 𝑣𝑦(𝑑), the π‘₯- and 𝑦-components of the velocity of the squirrel, as functions of time? (b) At 𝑑 = 5.00 s, how far is the squirrel from its initial position? (c) At 𝑑 = 5.00 s, what are the magnitude and direction of the squirrel’s velocity?


1
Expert's answer
2021-09-28T16:55:33-0400

(a) vx(t)=310.280+0.072tv_x(t)=310.280+0.072t


vy(t)=0.057t2v_y(t)=0.057t^2


(b) l=(310.280β‹…5+0.036β‹…52)2+(0.019β‹…53)2=1552.3 (m)l=\sqrt{(310.280\cdot5+0.036\cdot5^2)^2+(0.019\cdot5^3)^2}=1552.3\ (m)


(c) v=(310.280+0.072β‹…5)2+(0.057β‹…52)2=311 (m/s)v=\sqrt{(310.280+0.072\cdot5)^2+(0.057\cdot5^2)^2}=311\ (m/s)


tan⁑α=vyvx=1.425310.64=0.0046β†’Ξ±=0.26Β°\tan\alpha=\frac{v_y}{v_x}=\frac{1.425}{310.64}=0.0046\to \alpha=0.26Β° (above the x-axis)




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